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Given, loga\(x\) = \(\frac{1}{α}\), logb\(x\) = \(\frac{1}{β}\), logc\(x\) = \(\frac{1}{γ}\), then logabc\(x\) equals :

(a) αβγ 

(b) \(\frac{1}{αβγ}\)

(c) α + β + γ

(d) \(\frac{1}{α + β+γ}\)

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(d) \(\frac{1}{α + β+γ}\)

α = \(\frac{1}{log_ax}\), β = \(\frac{1}{log_bx}\), γ = \(\frac{1}{log_cx}\)

⇒ α = logxa, β = logxb, γ = logx

⇒ α + β + γ = logxa + logxb + logxc = logx(abc)

⇒ \(\frac{1}{α + β+γ}\) = \(\frac{1}{log_x(abc)}\) = logabcx.

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